Question

Three resistors $2\Omega ,4\Omega$ and $5\Omega$ are combined in parallel. This combination is connected to a battery of emf $20$V and negligible internal resistance. The total current drawn from the battery is?

• A. $10$A
• B. $15$A
• C. $19$A
• D. $23$A

Answer 1

Answer: (C)

$19$A

Since the three resistances are connected in parallel to each other, the net resistance can be obtained as:

$\frac{1}{R_{net}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

$\frac{1}{R_{net}}=\frac{1}{2}+\frac{1}{4}+\frac{1}{5}$

$R_{net}=\frac{2\times 4\times 5}{2\times 4+4\times 5+5\times 2}=\frac{2\times 4\times 5}{8+20+10}=\frac{2\times 4\times 5}{38}=\frac{20}{19}$

So, the current drawn from the battery is:

$I=\frac{V}{R}=\frac{20}{20}\times 19=19A$